Q.Draw qualitatively energy-level diagram showing d-orbital splitting in the octahedral environment. Predict the number of unpaired electrons in the complex [Fe(CN)6]4- . Is the complex diamagnetic or paramagnetic? Is it coloured? Explain.
Step 1. Oxidation state and configuration. [Fe(CN)6]4- has complex charge -4 and six CN- ligands (-1 each = -6): O.S.(Fe) . Fe2+ (Z=26) has a free-ion 3d6 configuration.
Step 2. Draw the octahedral CFT diagram (section 9.9.6, Fig. 9.2). The five degenerate d orbitals split into a lower t2g set (3 orbitals, lowered by 2/5 Delta-o) and an upper eg set (2 orbitals, raised by 3/5 Delta-o), separated by Delta-o.
Step 3. Decide high spin vs. low spin. CN- is a strong-field ligand (Table 9.6), producing a large Delta-o, so this is a LOW SPIN complex: the six 3d electrons pair up completely within the three t2g orbitals (t2g6 eg0, per Table 9.7's low-spin d6 row) rather than spreading into eg.
Step 4. Count unpaired electrons and magnetism. With t2g6 eg0, all six electrons are paired -- 0 unpaired electrons, so [Fe(CN)6]4- is diamagnetic.
Step 5. Colour. Even though the ground state has a fully-paired t2g6 configuration, a d-d transition promoting one electron from the filled t2g set up into the empty eg set (t2g6 -> t2g5eg1) is still possible on absorbing a photon of energy Delta-o (exactly the same mechanism as the [Ti(H2O)6]3+ example, section 9.9.8) -- so the complex IS coloured (in reality, [Fe(CN)6]4- is pale yellow), even though it is diamagnetic; being diamagnetic and being colourless are not the same thing.
0 unpaired electrons, diamagnetic; still coloured, via a t2g->eg d-d transition.
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